Introduction to Inequalities Roman Kvasov
Idea #1 :
Perform equivalent transformations of the initial inequality to bring it to the inequality that is obvious or known to be true
Problem Prove the inequality: a 2 + b 2 ≥ 2ab
Solution Start with the initial inequality and perform equivalent transformations passing all terms to one side and completing the square of the obtained expression: 2 a 2 + b 2 ≥ 2ab ⇔ a 2 + b 2 − 2ab ≥ 0 ⇔ ( a − b )( a − b ) ≥ 0 ⇔ ( a − b ) ≥ 0 Since the last inequality is obviously true then the initial inequality is true as well. 2 Note that the equality is reached only when ( a − b ) = 0 or equivalently when a = b .
Practice Problems 1. a 2 + 9b 2 ≥ 6ab , for any a, b 2. 2a 2 + b 2 + c 2 ≥ 2a ( b + c ) , for any a, b, c 3. a 2 + b 2 ≥ 2 ( a + b − 1) , for any a, b
4. a 2 + 6ab + 10b 2 ≥ 0 , for any a, b 5. a + b 2
2
( a + b) ≥
2
, for any a, b 2 1 1 6. a 2 + 2 ≥ a + , for any a a a
7. ( a 2 − b 2 ) ≥ 4ab ( a − b ) , for any a, b 2
2
8. ( a 3 − b3 ) ( a − b ) ≥ 3ab ( a − b ) , for any a, b 2
9. a 4 + b 4 ≥ a 3b + b3a , for any a, b
10. ( a 2 + b 2 )( c 2 + d 2 ) ≥ ( ac + bd ) , for any a, b, c, d 2
11.
2
≤ ab ≤
a+b a 2 + b2 ≤ , for a, b > 0 2 2
1 1 + a b 12. ( a + c )( b + d ) ≥ ab + cd , for any a, b, c, d > 0
Apply the AM-GM Inequality to the terms of the initial inequality
Idea #2 :
Problem Prove the inequality:
a b + ≥ 2 , for a, b > 0 b a
Solution Apply the AM-GM Inequality to the numbers a b + b a ≥ a⋅b 2 b a
a b and : b a
a b + b a≥ 1 ⇔ 2
a b + ≥2 b a a b Note that the equality is reached only when = or equivalently when a = b . b a ⇔
Practice Problems 1 1 1. ( a + b ) + ≥ 4 , for a, b > 0 a b 3 2. ( a + b )( a + b3 ) ≥ 4a 2b 2 , for a, b > 0
3. ( a + b )( ab + 9 ) ≥ 12ab , for a, b > 0
4. b ( a 2 + 1) + a ( b 2 + 1) ≥ 4ab , for a, b > 0
1 1 + 2 ≥ 2 , for a, b > 0 a +1 b +1 a 2 b2 c2 6. 1 + 1 + 1 + ≥ 8 , for a, b, c > 0 bc ac ab 5. a 2 + b 2 +
2
7. a 4 + b 4 + 2c 2 ≥ 4abc , for a, b, c > 0 a b c 8. + + ≥ 3 , for a, b, c > 0 b c a a+b b+c a+c 9. + + ≥ 6 , for a, b, c > 0 c a b 10. ( a + b + c ) ( a 2 + b 2 + c 2 ) ≥ 9abc , for a, b, c > 0 bc cd ad ab 11. 1 + 1 + 1 + 1 + ≥ 16 , for a, b, c, d > 0 ad ab bc cd 12. a + b + c ≥ ab + bc + ac , for a, b, c > 0 13. a 3 + b3 + c3 ≥ a 2 bc + b 2 ca + c 2 ab , for a, b, c > 0 14. a 4 + b 4 + c 4 ≥ abc ( a + b + c ) , for a, b, c > 0